Difference between revisions of "2015 AIME I Problems/Problem 1"
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==Video Solution For Problems 1-3== | ==Video Solution For Problems 1-3== | ||
https://www.youtube.com/watch?v=Vhlo_2zx2NY | https://www.youtube.com/watch?v=Vhlo_2zx2NY | ||
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==Solution 1== | ==Solution 1== | ||
We have <cmath>|A-B|=|1+3(4-2)+5(6-4)+ \cdots + 37(38-36)-39(1-38)|</cmath><cmath>\implies |2(1+3+5+7+ \cdots +37)-1-39(37)|</cmath><cmath>\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\implies \boxed{722}</cmath> | We have <cmath>|A-B|=|1+3(4-2)+5(6-4)+ \cdots + 37(38-36)-39(1-38)|</cmath><cmath>\implies |2(1+3+5+7+ \cdots +37)-1-39(37)|</cmath><cmath>\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\implies \boxed{722}</cmath> |
Revision as of 23:03, 8 January 2023
Contents
[hide]Problem
The expressions = and = are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers and .
Video Solution For Problems 1-3
https://www.youtube.com/watch?v=Vhlo_2zx2NY
Solution 1
We have
Solution 2
We see that
and
.
Therefore,
Solution 3 (slower solution)
For those that aren't shrewd enough to recognize the above, we may use Newton's Little Formula to semi-bash the equations.
We write down the pairs of numbers after multiplication and solve each layer:
and
Then we use Newton's Little Formula for the sum of terms in a sequence.
Notice that there are terms in each sequence, plus the tails of and on the first and second equations, respectively.
So,
Subtracting from gives:
Which unsurprisingly gives us
-jackshi2006
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.